Question

In the setup shown, a container is divided into eight cells and contains two molecules. Initially, both molecules are confined to the left side of the container. (a) Determine the number of possible arrangements before and after removal of the central barrier. (b) After the removal of the barrier, how many of the arrangements correspond to the state in which both molecules are in the left side of the container? How many correspond to the state in which both molecules are in the right side of the container? How many correspond to the state in which the molecules are in opposite sides of the container? Calculate the entropy for each state and comment on the most probable state of the system after removal of the barrier.

Short answer

After removing the barrier, the most probable state is the one where the molecules are on opposite sides. This state has the highest entropy.

Step-by-step solution

Determine Possible Arrangements Before Removing the Barrier

Since both molecules are initially confined to the left side of the container, and this left side consists of four cells, we need to find the number of ways to arrange the particles within these four cells. For 2 molecules in 4 cells, distinguishable molecules have \(4 \times 4 = 16\) arrangements if repetition where? Repetition allowed?

Determine Possible Arrangements After Removing the Barrier

With the barrier removed, the molecules can now occupy any of the 8 cells. Therefore, for 2 distinguishable molecules in 8 cells, the total arrangements are \(8 \times 8 = 64\) if repetition allowed.

Arrangements with Both Molecules on the Left Side

Both molecules in 4 left cells give us: \(4 \times 4 = 16\) arrangements.

Arrangements with Both Molecules on the Right Side

For both molecules to be on the right side in 4 cells, the number of possible arrangements is also \(4 \times 4 = 16\).

Arrangements with Molecules on Opposite Sides

There are two possibilities for this: one molecule on the left (4 choices) and one on the right (4 choices), resulting in \(4 \times 4 = 16\) arrangements. Another molecule swap gives \(4 \times 4 = 16\). Add both gives \( 16 + 16 = 32\)

Calculate Entropy for Each State

The entropy can be calculated using the formula \( S = k \ln \, (W) \), where \(k\) is the Boltzmann constant and \(W\) is the number of arrangements. For arrangements with both on the left or the right, \(W = 16\) and for opposite sides, \(W = 32\). \( S_{left,right} = k \ln \, (16) \), \( S_{opposite} = k \ln \, (32) \).

Determine the Most Probable State

The entropy directly relates to the probability of a state. Therefore, the most probable state corresponds to the greatest entropy. The state where the molecules are on opposite sides has the highest entropy \(S_{opposite}\), making it the most probable state after removing the barrier.

Key concepts

Molecular arrangements

Molecular arrangements are a way to describe the different ways molecules can be positioned within a container. In our scenario, we have two molecules and a container divided into eight cells. Initially, both molecules are on the left side, within four cells.
Without any barrier, molecules are free to occupy any available cell inside the container, creating multiple arrangements.

• When the center barrier is present, the arrangements are limited because both molecules are confined to the left side, leading to fewer positions for molecular arrangements.
• Removing the barrier allows molecules to spread into all eight cells, increasing the number of possible arrangements.

The key takeaway here is that the more space molecules have to roam, the more arrangements or states the system can achieve.

Boltzmann constant

The Boltzmann constant is a fundamental concept in thermodynamics and statistical mechanics. It is represented by the symbol \( k \) and is pivotal in calculations involving entropy.

1. Its role is to bridge the micro world of individual particles and the macro world of bulk properties by relating the energy of particles to absolute temperature.
2. The value of the Boltzmann constant is approximately \( 1.38 \times 10^{-23} \, J/K \).

In our exercise, the Boltzmann constant helps calculate the entropy associated with the different molecular arrangements. It serves as a linking factor between the number of states a system can have and the resulting entropy.

Probability of states

In statistical mechanics, the probability of states refers to the chance that a system will be found in a particular arrangement. When considering molecular arrangements, not all arrangements are equally probable.

• With the barrier removed, molecules can be anywhere, creating three distinct types of states: both molecules on the left, both on the right, or one in each half of the container.
• The more arrangements there are for a state, the higher its probability.

For instance, the state where each molecule is on opposite sides of the container has more arrangements, thus a higher probability after the barrier is removed. This state is considered the most probable due to its higher number of possible configurations.

Entropy calculation

Entropy is a measure of the disorder or randomness of a system. The connection between entropy and molecular arrangements is expressed mathematically through the formula \( S = k \ln(W) \) where \( S \) is entropy, \( k \) is the Boltzmann constant, and \( W \) is the number of possible arrangements.

• Greater \( W \) means greater entropy, implying the system is less ordered.
• In our problem scenario, we calculated entropy for various states: arrangements with molecules on the same side and those with them on opposite sides.
• The more arrangements available, as in the opposite side state, result in higher entropy, which signifies more disorder and higher probability for that state to occur.

Therefore, understanding entropy in terms of molecular arrangements and using the Boltzmann formula allows us to predict which state is more likely after changes, such as barrier removal, occur.